Respuesta :
Answer:
Probability that the sample will have a mean that is greater than $52,000 is 0.0057.
Step-by-step explanation:
We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.
We select a random sample of 1,000 people.
Let [tex]\bar X[/tex] = sample mean
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = $50,000
[tex]\sigma[/tex] = population standard deviation = $25,000
n = sample of people = 1,000
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
So, probability that the sample will have a mean that is greater than $52,000 is given by = P([tex]\bar X[/tex] > $52,000)
P([tex]\bar X[/tex] > $52,000) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } }[/tex] ) = P(Z > 2.53) = 1 - P(Z [tex]\leq[/tex] 2.53)
= 1 - 0.9943 = 0.0057
Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.
Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.
